In this section we'll discuss motion on an incline or a hill.  We will not yet be considering air drag.  For now we will only consider the gravitational force and friction between the hill and the object.  Real-life scenarios might include skating down a hill, skiing or snowboarding, or riding a bike down a hill.  Just realize that if speed gets high in any of those scenarios that air drag is an important factor.

If you look at the diagram above, you can see the effect that the incline has on the forces including friction.  The first thing to note is that[b] the coordinate system has been chosen so that one axis (x-axis) is parallel to the acceleration,[/b] which will be down the incline.  Therefore the gravitational force needs to be broken into components along those coordinate axes.  In contrast with that, the normal force, which is always normal (perpendicular) to the incline, is in the +y direction and doesn't need to be broken into components.  Notice that since only a part of the gravitational force is pressing the block against the incline (and the rest down the incline), that the normal force is smaller in this case than it would be on a level surface.   [br][br]For a static block, the friction force will never be larger than F[sub]g,x[/sub], since that's the force trying to make the block slide and friction tries to prevent the slide.  In this sense it only needs to counter F[sub]g,x[/sub]. So regardless of how you change the friction coefficient, the friction force is limited in magnitude by F[sub]g,x[/sub].  If you make the coefficient smaller until the friction force starts shrinking rather than being limited by F[sub]g,x[/sub], the block will slide down the incline.  This is because there is a net force down the incline given by: [br][br][center][math]\Sigma F_x=F_{g,x}-F_f = mg\sin\theta - \mu mg\cos\theta = ma_x.[/math][/center] [br]Please note that these terms are all scalars since we're adding x-components of vectors, so for instance the term F[sub]f[/sub] is the magnitude of the friction force. The negative indicates that the force acts up the incline in the -x direction. In another form we could write [math]\vec{F}_f=-F_f\;\hat{i}+0\hat{j}.[/math][br][br]Another thing to note is that [b]a frictionless block will slide down an incline with an acceleration given by [math]a=g\sin\theta.[/math][/b]  This can be seen from the sum of the forces in the x-direction equation above.  Without friction we just have one term on the left side, and the resulting acceleration is obvious.  It's always a good idea to analyze a result so that you get the "message" that it's trying to convey and also potentially find errors in your algebra.  In this case for instance, what happens when the angle gets large?  Do you see that a=g when the slope becomes a cliff (since the sine of 90 degrees is one)?  Also, when the slope is flat a=0 (since the sine of zero is zero).  I hope these both make sense to you.

As we just discussed, the component of gravity that causes an object to descend down a hill or a ramp is [math]F=mgsin\left(\Theta\right).[/math] It turns out that when the angle of incline is small, we can make a very simple and useful approximation for this force. Mathematically, if you compare the sine of a small angle with the tangent of the same small angle, they are virtually the same. This is because tan=sin/cos, and cosine is nearly 1 for small angles. [br][br]Since road surfaces can almost always be considered small angles, we can replace the sine in the force equation with the tangent for many realistic scenarios. This is useful because the tangent is just the grade (often listed as a percentage) of the road surface. What this means is that when I'm rolling down a 5% grade, if I take 5% of the force of gravity on me, that's the force causing me to descend the hill. So F=5%(mg)=0.05mg. [br][br]How accurate is the approximation? Even up to 20% grades (very, very steep for a road) it only gives 6% error. At more typical slopes like a 7% grade, which is the largest allowed slope on a freeway in the US, the error is only 0.75%.

[color=#1e84cc]Q: How large is the force of gravity that pulls a 2000kg car down a 7% freeway descent in Colorado?[br]A: F=mg(grade) = 2000kg(10N/kg)(0.07) = 140N. Simple, right?[/color]

[i][color=#1e84cc]Imagine an incline is slowly raised to a steeper and steeper angle. (Don't use that small angle approximation here.) To find the angle at which a block with a given friction coefficient would slip, we only need to compare [/color][math]F_{g,x}[/math][color=#1e84cc] and [/color][math]F_{f}.[/math][color=#1e84cc] It is not permissible in this problem for friction to be the larger term because that would imply that the block would start accelerating up the incline due to the force of friction! So the only two possibilities are that friction is smaller than or equal to [/color][math]F_{g,x}.[/math][color=#1e84cc] [br]When it becomes smaller than [/color][math]F_{g,x}[/math][color=#1e84cc], that's when the block would slip. So the block would slide if [/color][math]F_f[color=#1e84cc], or if [/color][math]\mu mg\cos\theta[color=#1e84cc] This clearly leads to the condition that when [/color][math]\tan\theta>\mu[/math][color=#1e84cc] the block slides down the incline. A similar result to the problem in the previous section, but for a different scenario. [/color][/i]

When skating on a half pipe without doing anything to keep your speed up, you will eventually come to rest.  That is to say that you lose speed more rapidly going up the ramp than you gain it going down the ramp.  The effect of this over a few runs back and forth is that you slow down dramatically and don't rise as high on the ramp.  [br][br]Let's look at calculating the rate of acceleration going up a ramp versus going down a ramp.  At first inspection it may look like the forces are identical.  If you look carefully, one of the forces changes.  Can you look at the force diagram above and see which one changes?  Think about this for a moment.[br][br]It turns out that the friction force is not always directed up the ramp.  Instead, it always opposes the motion.  When the skater is going up the ramp the friction acts down the ramp and when the skater goes down the ramp the friction goes up the ramp.  Gravity, however is the same both ways.  Therefore when the skater is going up the ramp, both friction and gravity work together to slow the skater.  When going down the ramp friction fights gravity.  The friction is trying to slow the skater while gravity is trying to speed them up.  Let's consider the +x direction to be down the ramp.  When the skater is going down the ramp we get [math]\Sigma F_x=mg\sin\theta-\mu mg\cos\theta = ma_{x,down}.[/math]  When the skater is going up the ramp we get [math]\Sigma F_x=mg\sin\theta+\mu mg\cos\theta = ma_{x,up}.[/math] [br][br]If we use some numbers such as [math]\mu = 0.05[/math] and [math]\theta = 5^o,[/math] and assume g=10m/s[sup]2[/sup] we would get [math]a_{x,up}= 1.37m/s^2[/math] and [math]a_{x,down}= 0.37m/s^2.[/math] It is clear on this modest slope that friction dramatically changes the upward versus the downward acceleration![br][br][b]Finding Slope and Friction Coefficient from Acceleration[/b][br]There are two other interesting uses of this physics. If you had a means of measuring acceleration (perhaps with video frames that are time-stamped, or with a motion sensor), you could determine the slope of a ramp by only measuring acceleration - even though friction is present. You could also determine the coefficient of friction this way... and you wouldn't need a force probe or anything. Look at the two equations for going up and going down the ramp with the mass terms cancelled: [br][br][center][math]g\sin\theta-\mu g\cos\theta = a_{x,down}\\[br]g\sin\theta+\mu g\cos\theta = a_{x,up}.[/math][/center][br]If you add the two equations together you can find the slope of the ramp, and by taking the difference you can find the coefficient of friction. I'll leave it to you to work out the details of the algebra.